Answer:
Saturated solution
We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.
Explanation:
Step 1: Calculate the mass of water
The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.
Step 2: Calculate the mass of glucose per 100 g of water
550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.
Step 3: Classify the solution
The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.
Answer:
The question isn't worded properly, but if 1 or 2 are DECREASED, the frequency of collisions of specified molecules will decrease.
Explanation:
Catalysts only facilitate reaction once molecules collide. Increased temperature makes molecules move more, and thus collide more. For concentration, if there are more molecules in the same amount of room/liquid, there will be more collisions because there are more of the molecules to collide.
Answer:
Explanation:
<u>1) Data:</u>
a) V = 93.90 ml
b) T = 28°C
c) P₁ = 744 mmHg
d) P₂ = 28.25 mmHg
d) n = ?
<u>2) Conversion of units</u>
a) V = 93.90 ml × 1.000 liter / 1,000 ml = 0.09390 liter
b) T = 28°C = 28 + 273.15 K = 301.15 K
c) P₁ = 744 mmHg × 1 atm / 760 mmHg = 0.9789 atm
d) P₂ = 28.5 mmHg × 1 atm / 760 mmHg = 0.0375 atm
<u>3) Chemical principles and formulae</u>
a) The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Hence, the partical pressure of the hydrogen gas collected is equal to the total pressure less the vapor pressure of water.
b) Ideal gas equation: pV = nRT
<u>4) Solution:</u>
a) Partial pressure of hydrogen gas: 0.9789 atm - 0.0375 atm = 0.9414 atm
b) Moles of hygrogen gas:
pV = nRT ⇒ n = pV / (RT) =
n = (0.9414 atm × 0.09390 liter) / (0.0821 atm-liter /K-mol × 301.15K) =
n = 0.00358 mol (which is rounded to 3 significant figures) ← answer
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